Simple ODE with given initial value Find all solutions of $$2x'=3x^{1/3}, \ x(0)=0,\ x(1)=a$$
for which $\lim_{t\rightarrow -\infty}x(t) = 0 $, in respect to $a$ parameter, $a\in\mathbb{R}$.
It's quite easy to notice that the solutions are $x(t)=t^{3/2}, \ x(t)=-t^{3/2}$ and$ \ x(t)=0$. Does that mean the only solution that meets this exercise requirements is $x(t)\equiv0$ along with $a=0$?
 A: The function $x^{2/3}$ is not Lipschitz at $x=0$. This means that uniqueness of solution may fall, as in fact it does. Given $t_-\le0\le t_+$, the function
$$
x(t)=\begin{cases}
-(t_--t)^{3/2} & \text{if }t<t_-,\\
0 & \text{if } t_-\le t\le t_+,\\
(t-t_+)^{3/2} & \text{if }t>t_+,
\end{cases}
$$
is a solution. From here you should be able to show that there is a unique solution if $0<a\le1$, no solution if $a>1$ or $a<0$ and infinite solutions if $a=0$.
A: General solution: $x=\sqrt{t+k}$. We want $x(0)=0$, so $k=0$. That means we must have $x(1)=1$ and hence $a=1$. It also means we have a difficulty with $t<0$. In fact, there is a difficulty at $t=0$. Because $x'(0)=\infty\ne 1.5x^{1/3}$. 
Note that since we are stuck with a derivative discontinuity at $t=0$ in any case we could get a solution for negative $t$ by taking $x(t)=0$ for all negative $t$. (It would be continuous at $t=0$, but the derivative would be discontinuous there.)  
Of course, once you accept that solutions (which are differentiable everywhere) are impossible for $\textit{any}$ $a$ except $a=0$ (where the solution $x(t)=0$ for all $t$ is fine), there is no reason to restrict yourself to one derivative discontinuity. Aguirre's solution shows how you could patch three pieces together to get $x(0)=0,x(1)=a$ for $0<a\le1$. The reason it will not work for $a>1$ is that you cannot vary $k$ or the starting position $t_+$ (in Aguirre's solution) to get the $x(t)$ to grow fast enough.
